Identify the organic functional group present in both vegetable oil and biodiesel.

For part of her extended essay investigation into the efficiency of the process, a student reacted a pure sample of a vegetable oil (where \({\text{R}} = {{\text{C}}_{{\text{17}}}}{{\text{H}}_{{\text{33}}}}\)) with methanol. The raw data recorded for the reaction is below.

\[\begin{array}{*{20}{l}} {{\text{Mass of oil}}}&{ = 1013.0{\text{ g}}} \\ {{\text{Mass of methanol}}}&{ = 200.0{\text{ g}}} \\ {{\text{Mass of sodium hydroxide}}}&{ = 3.5{\text{ g}}} \\ {{\text{Mass of biodiesel produced}}}&{ = 811.0{\text{ g}}} \end{array}\]

The relative molecular mass of the oil used by the student is 885.6. Calculate the amount (in moles) of the oil and the methanol used, and hence the amount (in moles) of excess methanol.

State what is meant by the term dynamic equilibrium.

Using the abbreviations [vegetable oil], [methanol], [glycerol] and [biodiesel] deduce the equilibrium constant expression \({\text{(}}{K_{\text{c}}}{\text{)}}\) for this reaction.

Suggest a reason why excess methanol is used in this process.

State and explain the effect that the addition of the sodium hydroxide catalyst will have on the position of equilibrium.

The reactants had to be stirred vigorously because they formed two distinct layers in the reaction vessel. Explain why they form two distinct layers and why stirring increases the rate of reaction.

Calculate the percentage yield of biodiesel obtained in this process.

ester;

amount of oil \( = \frac{{1013.0}}{{885.6}} = 1.144{\text{ mol}}\);

amount of methanol \( = \frac{{200.0}}{{32.05}} = 6.240{\text{ mol}}\);

since three mol of methanol react with one mol of vegetable oil the amount of excess methanol \( = 6.204 - (3 \times 1.144) = 2.808{\text{ mol}}\);

rate of the forward reaction is equal to the rate of the reverse reaction / forward and reverse reactions occur and the concentrations of the reactants and products do not change / OWTTE;

\({K_{\text{c}}} = \frac{{{\text{[glycerol]}} \times {{{\text{[biodiesel]}}}^3}}}{{{\text{[vegetable oil]}} \times {{{\text{[methanol]}}}^3}}}\);

to move the position of equilibrium to the right/product side / increase the yield of biodiesel;

no effect (on position of equilibrium);

increases the rate of the forward and the reverse reactions equally (so equilibrium reached quicker) / it lowers Ea for both the forward and reverse reactions by the same amount / OWTTE;

No ECF for explanation.

vegetable oil is mainly non-polar and methanol is polar / OWTTE;

stirring brings them into more contact with each other / increase the frequency of collisions / OWTTE;

Do not allow simply mixing.

(relative molecular mass of biodiesel, \({{\text{C}}_{19}}{{\text{H}}_{36}}{{\text{O}}_2} = 296.55\))

maximum yield of biodiesel \( = 3.432{\text{ mol}}/1018{\text{ g}}\);

percentage yield \(\frac{{811.0}}{{1018}} \times 100 = 79.67\% \);

Allow 80% for percentage yield.

Part (a) was reasonably well answered with most candidates opting for an ester. Ketone (frequently spelt keytone) and carbonyl were the most common incorrect responses.

In Part (b) most candidates scored 1 or 2 marks, showing that they knew the correct method but the third mark proved to be more difficult to obtain, usually because the factor of 3 was omitted.

In general, equilibrium (Part (c)) seems to be quite well understood. The most common error in (i) was to describe the reaction as constant rather than having opposing reactions with equal rates.

The expression in (ii) was an easy mark for the better candidates. The weaker ones often missed one or both of the powers of three and a small number had + signs in both the numerator and denominator.

In (iii) the most common incorrect answer was ‘to use up all the vegetable oil’.

In (iv) most candidates were aware that a catalyst has no effect on the equilibrium constant but failed to gain the second mark for saying that the catalyst affected both reactions equally, either by increasing the rates equally or lowering the activation energy by the same amount.

Very few candidates scored both marks for Part (d) of the question. The better candidates realised that there was a difference in polarity, though not always identifying which reactant was polar and which was non-polar. The most common answers either simply stated that the two were immiscible or that they had different densities. For the second mark an increase in collisions was often mentioned but not always an increase in the frequency of collisions.

Candidates found Part (e) to be very difficult. This was not helped by the small amount of space available to them on the paper. Many answers expressed the data in terms which would have calculated (100 - %) as though they had been drilled to calculate % impurities.